(n - 10)(n + 4)
-n(n + 10) or n(-n - 10)
5(n2 + 2n + 4)
Take 5 out. If the missing signs are pluses, it becomes 5(n2 + 2n + 4) If the missing signs are minuses, it becomes 5(n2 - 2n - 4)
n2-1 and n2-4 are trivial cases because of n2-m2=(n-m)(n+m). So the only prime of the form n2-1 is 3 and of the form n2-4 is 5.
-mn(m - n2)(m + n2)
n2 + 9n + 18 = (n + 6)(n + 3)
If you mean: n2+16n+64 then it is (n+8)(n+8) when factored
n2 + 7n - 44 (n + 11)(n - 4)
(n - 10)(n + 4)
(n + 11)(n - 4)
(n + 4)(n - 4)
-n(n + 10) or n(-n - 10)
This can not be solved as it is not an equation but a simple expression. If you are looking to factor it, then the answer is: n2 - 3n - 18 = (n - 6)(n + 3)
5(n2 + 2n + 4)
square n x n = n2
n2 - 100 = (n + 10)(n - 10)